Let $$P(x)$$ be a polynomial of degree n with leading coefficient $$1$$ . Let $$v(x)$$ be any function and $$v1(x)=$$∫$$v(x)dxv2(x)=$$∫$$v1(x)dx$$…$$vn+1(x)=$$∫$$vn(x)dx$$ then ∫$$P(x)v(x)dx$$ is equal to

Correct option is 2P(x)v1(x)−P′(x)v2(x)+P′′(x)v3(x)…+(−1)nn!vn+1(x)

Questions

Let $P (x)$ be a polynomial of degree $n$ with leading
coefficient 1 . Let $v (x)$ be any function and $v_{1} (x) = \int v (x) d x$
$v_{2} (x) = \int v_{1} (x) d x \dots v_{n + 1} (x) = \int v_{n} (x) d x$ then $\int P (x) v (x) d x$ is equal to

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P(x)v1(x)+P′(x)v2(x)2!+P′′(x)v3(x)3!…+vn+1(x)(n+1)!

P(x)v1(x)−P′(x)v2(x)+P′′(x)v3(x)…+(−1)nn!vn+1(x)

P(x)v1(x)+P′(x)v2(x)+P′′(x)v3(x)…+nvn+1(x)

P(x)v1(x)−P′(x)v2(x)2!+P′′(x)v3(x)3!…+(−1)nvn+1(x)(n+1)!

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detailed solution

Correct option is B

Integrating by part repeatedly , we get ∫P(x)v(x)dx=P(x)v1(x)−∫P′(x)v1(x)dx=P(x)v1(x)−P′(x)v2(x)+∫P′′(x)v2(x)dx=P(x)v1(x)−P′(x)v2(x)+P′′(x)v3(x)−….+(−1)n∫P(n)(x)vn(x)dxNow , use P(n)(x)=n!

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Let P(x) be a polynomial of degree n with leading coefficient 1 . Let v(x) be any function and v1(x)=∫v(x)dxv2(x)=∫v1(x)dx…vn+1(x)=∫vn(x)dx then ∫P(x)v(x)dx is equal to

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Ready to Test Your Skills?

free study material

Let $P (x)$ be a polynomial of degree $n$ with leading
coefficient 1 . Let $v (x)$ be any function and $v_{1} (x) = \int v (x) d x$
$v_{2} (x) = \int v_{1} (x) d x \dots v_{n + 1} (x) = \int v_{n} (x) d x$ then $\int P (x) v (x) d x$ is equal to